Editing Tangent half-angle formula (section)

===Algebraic proofs===
Using the angle addition and subtraction formulae for both the sine and cosine one obtains
<math display="block">
\begin{align}
\sin (a+b) + \sin (a-b) &= 2 \sin a \cos b \\[15mu]
\cos (a+b) + \cos (a-b) & = 2 \cos a \cos b\,.
\end{align}
</math>

Setting <math display="inline">a= \tfrac12 (\eta+\theta)</math> and <math>b= \tfrac12 (\eta-\theta)</math> and substituting yields
<math display="block">
\begin{align}
\sin \eta + \sin \theta = 2 \sin \tfrac12(\eta+\theta) \, \cos \tfrac12(\eta-\theta) \\[15mu]
\cos \eta + \cos \theta = 2 \cos\tfrac12(\eta+\theta) \, \cos\tfrac12(\eta-\theta)\,.
\end{align}
</math>

Dividing the sum of sines by the sum of cosines gives
<math display="block">\frac{\sin \eta + \sin \theta}{\cos \eta + \cos \theta} = \tan \tfrac12(\eta+\theta)\,.</math>

Also, a similar calculation starting with <math>\sin (a+b) - \sin (a-b)</math> and <math>\cos (a+b) - \cos (a-b)</math> gives
<math display="block">-\frac{\cos \eta - \cos \theta}{\sin \eta - \sin \theta} = \tan \tfrac12(\eta+\theta)\,.</math>

Furthermore, using [[double-angle formulae]] and the Pythagorean identity <math display="inline">1 + \tan^2 \alpha = 1 \big/ \cos^2 \alpha</math> gives
<math display="block">
\sin \alpha
= 2\sin \tfrac12 \alpha \cos \tfrac12 \alpha
= \frac{ 2 \sin \tfrac12 \alpha\, \cos \tfrac12 \alpha
           \Big/ \cos^2 \tfrac12 \alpha}
       {1 + \tan^2 \tfrac12 \alpha}
= \frac{2\tan \tfrac12 \alpha}{1 + \tan^2 \tfrac12 \alpha}
</math>
<math display="block">
\cos \alpha
= \cos^2 \tfrac12 \alpha - \sin^2 \tfrac12 \alpha
= \frac{ \left(\cos^2 \tfrac12 \alpha - \sin^2 \tfrac12 \alpha\right)
           \Big/ \cos^2 \tfrac1 2 \alpha}
       {  1 + \tan^2 \tfrac12 \alpha}
= \frac{1 - \tan^2 \tfrac12 \alpha}{1 + \tan^2 \tfrac12 \alpha}\,.
</math>
Taking the quotient of the formulae for sine and cosine yields
<math display="block">\tan \alpha = \frac{2\tan \tfrac12 \alpha}{1 - \tan ^2 \tfrac12 \alpha}\,.</math>